Question

A double-slit apparatus is immersed in a liquid of refractive index $\mathrm{1.33.}$ It has slit separation of $1mm$ and distance between the plane of slits and screen $1.33m$. The slits are illuminated by a parallel beam of light whose wavelength in air is $6300\mathring{A}$. The fringe width is:

• A. $1.33$ mm
• B. $\frac{0.63}{1.33}$ mm
• C. $\frac{0.63}{{1.33}^2}$ mm
• D. $0.63$ mm

Answer 1

The correct option is D $0.63$ mm

Fringe width $f=\frac{\lambda D}{d}$
where:

$D$ is distance between the plane of slits and screen.
$d$ is slit separation.
$\lambda$ is wavelength.

In liquid $\lambda =\frac{6300\mathring{A}}{1.33}$

$f=\frac{6300\times {10}^{-7}mm\times 1.33m}{1.33\times {10}^{-3}m}$

$=0.63mm$